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olga2289 [7]
3 years ago
15

What minimum distance should you separate two sources emitting the same waves with wavelength 5mm in phase such that you obtain

complete destructive interference along the line connecting the sources and to the right of the sources?
Physics
1 answer:
Makovka662 [10]3 years ago
8 0

To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as

d = n \frac{\lambda}{2}

Where,

d = Path difference

\lambda= wavelength

n = Any integer which represent the number of repetition of the spectrum

In this question the distance between the two source will be minimum for the case of minimum path difference, then n= 1

d = \frac{\lambda}{2}

d = \frac{5*10^{-3}}{2}

d = 2.5mm

Therefore the minimum distance that should you separate two sources emitting the same waves is 2.5mm

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A car travelling at a constant speed of 70km/h passes a stationary police car. The police car immediately goes on the chase acce
Virty [35]

Answer:

18.24 seconds

Explanation:

First you convert the km/h to m/s, 70km/h=(175/9)m/s,85km/h=(425/18)m/s.

You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.

5 0
3 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

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